Đáp án:

a) \(x \in \emptyset \)

b) x=3

c) \(x = \dfrac{1}{3}\)

d) \(\left[ \begin{array}{l}
x = \dfrac{5}{2}\\
x =  - \dfrac{2}{3}
\end{array} \right.\)

e) \(x = \dfrac{3}{8}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)DK:x \ge 4\\
\sqrt {x + 3}  = 4 + \sqrt {x - 4} \\
 \to x + 3 = 16 + 8\sqrt {x - 4}  + x - 4\\
 \to 9 =  - 8\sqrt {x - 4} \left( {KTM} \right)\\
 \to x \in \emptyset \\
b)DK:x \ge 2\\
\sqrt {3x - 5}  + \sqrt {x - 2}  = 3\\
 \to 3x - 5 + 2\sqrt {\left( {3x - 5} \right)\left( {x - 2} \right)}  + x - 2 = 9\\
 \to 2\sqrt {3{x^2} - 11x + 10}  = 16 - 4x\\
 \to \sqrt {3{x^2} - 11x + 10}  = 8 - 2x\\
 \to 3{x^2} - 11x + 10 = 64 - 32x + 4{x^2}\left( {DK:4 \ge x} \right)\\
 \to {x^2} - 21x + 54 = 0\\
 \to \left( {x - 18} \right)\left( {x - 3} \right) = 0\\
 \to \left[ \begin{array}{l}
x = 18\left( l \right)\\
x = 3
\end{array} \right.\\
c)\sqrt {{{\left( {x - 2} \right)}^2}}  = 2x + 1\\
 \to \left| {x - 2} \right| = 2x + 1\\
 \to \left[ \begin{array}{l}
x - 2 = 2x + 1\left( {DK:x \ge 2} \right)\\
x - 2 =  - 2x - 1\left( {DK:x < 2} \right)
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x =  - 3\left( l \right)\\
3x = 1
\end{array} \right.\\
 \to x = \dfrac{1}{3}\\
d)\sqrt {{{\left( {5x - 3} \right)}^2}}  = x + 7\\
 \to \left| {5x - 3} \right| = x + 7\\
 \to \left[ \begin{array}{l}
5x - 3 = x + 7\\
5x - 3 =  - x - 7
\end{array} \right.\\
 \to \left[ \begin{array}{l}
4x = 10\\
6x =  - 4
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = \dfrac{5}{2}\\
x =  - \dfrac{2}{3}
\end{array} \right.\\
e)\sqrt {{{\left( {5x - 1} \right)}^2}}  = 3x - 2\\
 \to \left| {5x - 1} \right| = 3x - 2\\
 \to \left[ \begin{array}{l}
5x - 1 = 3x - 2\left( {DK:x \ge \dfrac{1}{5}} \right)\\
5x - 1 =  - 3x + 2\left( {DK:x < \dfrac{1}{5}} \right)
\end{array} \right.\\
 \to \left[ \begin{array}{l}
2x =  - 1\\
8x = 3
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x =  - \dfrac{1}{2}\left( l \right)\\
x = \dfrac{3}{8}
\end{array} \right.
\end{array}\)